What is the solution for y^{''}+3y^'-4y=2cos(x) ?

The solution for y^{''}+3y^'-4y=2cos(x) is y=c_{1}e^x+c_{2}e^{-4x}+3/17 sin(x)-5/17 cos(x)

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y^{''}+3y^'-4y=2cos(x)

y^{^{''}} + 3 y^{^{'}} - 4 y = 2 cos (x)

y = c_{1} e^{x} + c_{2} e^{- 4 x} + \frac{3}{17} sin (x) - \frac{5}{17} cos (x)

Solution steps

y^{''} (x) + 3 y^{'} (x) - 4 y = 2 cos (x)

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 4 x} + \frac{3}{17} sin (x) - \frac{5}{17} cos (x)

y = c_{1} e^{x} + c_{2} e^{- 4 x} + \frac{3}{17} sin (x) - \frac{5}{17} cos (x)

What is the solution for y^{''}+3y^'-4y=2cos(x) ?
The solution for y^{''}+3y^'-4y=2cos(x) is y=c_{1}e^x+c_{2}e^{-4x}+3/17 sin(x)-5/17 cos(x)